Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
vector<pair<int, int> > ans1, ans2;
int g[101][101], a[101][101];
int n, m;
void checkrow(vector<pair<int, int> >& ans) {
  for (int i = 0; i < n; i++) {
    int curmin = inf;
    for (int j = 0; j < m; j++) {
      curmin = min(g[i][j], curmin);
    }
    for (int j = 0; j < m; j++) {
      g[i][j] -= curmin;
    }
    if (curmin > 0)
      while (curmin--) ans.push_back({0, i + 1});
  }
}
void checkcoll(vector<pair<int, int> >& ans) {
  for (int j = 0; j < m; j++) {
    int curmin = inf;
    for (int i = 0; i < n; i++) {
      curmin = min(curmin, g[i][j]);
    }
    for (int i = 0; i < n; i++) {
      g[i][j] -= curmin;
    }
    if (curmin)
      while (curmin--) ans.push_back({1, j + 1});
  }
}
bool check(vector<pair<int, int> >& ans, bool way) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      g[i][j] = a[i][j];
    }
  }
  for (int i = 0; i < 100; i++) {
    if (way)
      checkcoll(ans);
    else
      checkrow(ans);
    way = way ^ 1;
  }
  bool ok = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      ok &= g[i][j] == 0;
    }
  }
  return ok;
}
void print(vector<pair<int, int> > ans) {
  cout << ans.size() << endl;
  for (auto x : ans) {
    cout << (x.first ? "col" : "row") << ' ' << x.second << endl;
  }
}
int main() {
  cin >> n;
  cin >> m;
  bool ok = 1;
  int curmin = 1e9 + 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j];
      g[i][j] = a[i][j];
    }
  }
  bool ok1 = check(ans1, 0);
  bool ok2 = check(ans2, 1);
  int minn = inf;
  if (!ok1)
    if (ok2) {
      print(ans2);
    } else
      cout << -1 << endl;
  else if (!ok2)
    if (ok1)
      print(ans1);
    else
      cout << -1 << endl;
  else if (ans1.size() > ans2.size())
    print(ans2);
  else
    print(ans1);
  return 0;
}